Inverse Functions
The inverse of any function $ \displaystyle (f)$ is a function that will do the opposite of $ \displaystyle f$. In other words the function that will undo the effects of $ \displaystyle f$. So, if $ \displaystyle f$ maps 4 into 13, then the inverse of $ \displaystyle f$ will map 13 onto 4.
In effect, when $ \displaystyle f$ is applied to a number and the reverse of $ \displaystyle f$ is applied to the result, you will get back to the number you started with.
In simple cases, you can find the inverse of a function by inspection. For example, the inverse of $ \displaystyle x\to x+5$ must be $ \displaystyle x\to x-5$ because subtraction is the inverse of addition, to undo add five you have to subtract five.
Similarly, the inverse of $ \displaystyle x\to 2x$ is $ \displaystyle x\to \frac{x}{2}$, because to undo multiply by two you have to divide by two.
The inverse of the function $ \displaystyle (f)$ is written as $ \displaystyle {{f}^{{-1}}}$
So if$ \displaystyle (f)=x+5$, then $ \displaystyle {{f}^{{-1}}}(x)=x-5$ and if $ \displaystyle g(x)=2x$ then$ \displaystyle {{g}^{{-1}}}(x)=\frac{x}{2}$
Some functions do not have an inverse. Think about the function $ \displaystyle x\to 2x$. This is a function because for every value of$ \displaystyle x$, there is only one value of $ \displaystyle {{x}^{2}}$. The inverse (in other words, the square root) is not a function because a positive number has two square roots, one negative, and one positive.
Finding the inverse of a function
There are two methods of finding the inverse:
Method 1: Using a flow diagram
In this method you draw a flow diagram for the function and then work out the inverse by “reversing” the flow to undo the operations in the boxes.
Method 2: Reversing the mapping
In this method you use the fact that if $ \displaystyle f$ maps $ \displaystyle x$ onto $ \displaystyle y$, then $ \displaystyle {{f}^{{-1}}}$maps$ \displaystyle y$ onto $ \displaystyle x$. To find $ \displaystyle {{f}^{{-1}}}$you have to find a value of $ \displaystyle x$ that corresponds to a given value of$ \displaystyle y$.
Let’s see some examples using both methods.
Example 1: Find the inverse of$ \displaystyle f(x)=3x-4$.
Using method 1, “the flow diagram” we get:
$ \displaystyle f:input\to \times 3\to -4\to output$
$ \displaystyle {{f}^{{-1}}}:output\leftarrow \div 3\leftarrow +4\leftarrow input$
Let $ \displaystyle x$ be the input to $ \displaystyle {{f}^{{-1}}}$
$\displaystyle \frac{{x+4}}{3}\leftarrow \div 3\leftarrow +4\leftarrow x$
$ \displaystyle {{f}^{{-1}}}(x)=\frac{{x+4}}{3}$
Using method 2, “reversing the mapping”
Suppose the function maps $ \displaystyle x$ onto $ \displaystyle y$ ( $ \displaystyle y$ is the subject). Make $ \displaystyle x$ the subject of the formula, so that $ \displaystyle y$ maps into $ \displaystyle x$.
$ \displaystyle y=3x-4$
$ \displaystyle y+4=3x$
$ \displaystyle x=\frac{{y+4}}{3}$
You know that $ \displaystyle {{f}^{{-1}}}$ maps $ \displaystyle y$ onto $ \displaystyle x$, so $ \displaystyle {{f}^{{-1}}}(y)=\frac{{y+4}}{3}$
This is usually written in terms of $ \displaystyle x$ so $ \displaystyle {{f}^{{-1}}}(x)=\frac{{x+4}}{3}$
Example 2: Given $ \displaystyle g(x)=5-2x$, find $ \displaystyle {{g}^{{-1}}}(x)$.
Using method 1, “the flow diagram” you get:
$ \displaystyle g:input\to \times (-2)\to +5\to output$
$ \displaystyle {{g}^{{-1}}}:output\leftarrow \div (-2)\leftarrow -5\leftarrow output$
Let $ \displaystyle x$ be the input to $ \displaystyle {{g}^{{-1}}}$
$ \displaystyle \frac{{x-5}}{{-2}}\leftarrow \div (-2)\leftarrow -5\leftarrow x$
$ \displaystyle {{g}^{{-1}}}(x)=\frac{{x-5}}{{-2}}=\frac{{5-x}}{2}$
Using method 2, “reversing the mapping”
This means $ \displaystyle g$ maps $ \displaystyle x$ onto $ \displaystyle y$
Make $ \displaystyle x$ the subject of the formula so, that $ \displaystyle y$ maps onto $ \displaystyle x$.
Let $ \displaystyle y=5-2x$
$ \displaystyle 2x=5-y$
$ \displaystyle x=\frac{{5-y}}{2}$
$ \displaystyle {{g}^{{-1}}}$ maps $ \displaystyle y$ onto$ \displaystyle x$, so $ \displaystyle {{g}^{{-1}}}(y)=\frac{{5-y}}{2}$
This is usually written in terms of $ \displaystyle x$ so, $ \displaystyle {{g}^{{-1}}}(x)=\frac{{5-x}}{2}$
Example 3: Given the function $ \displaystyle g(x)=\frac{x}{3}-44$ find $ \displaystyle {{g}^{{-1}}}(x)$.
Using method 1, “the flow diagram” you get:
$ \displaystyle g:input\to \div 3\to -44\to output $
$ \displaystyle {{g}^{{-1}}}:output\leftarrow \times 3\leftarrow +44\leftarrow output$
Let $ \displaystyle x$ be the input to $ \displaystyle {{g}^{{-1}}}$
$ \displaystyle 3(x+44)\leftarrow \times 3\leftarrow +44\leftarrow x$
$ \displaystyle {{g}^{{-1}}}(x)=3(x+44)=3x+132$
Using method 2, “reversing the mapping“
This means $ \displaystyle g$ maps $ \displaystyle x$ onto $ \displaystyle y$
Make $ \displaystyle x$ the subject of the formula so, that $ \displaystyle y$ maps onto $ \displaystyle x$.
Let $ \displaystyle y=\frac{x}{3}-44$
$ \displaystyle y+44=\frac{x}{3}$
$ \displaystyle x=3(y+44)$
$ \displaystyle {{g}^{{-1}}}$ maps $ \displaystyle y$ onto$ \displaystyle x$, so $ \displaystyle {{g}^{{-1}}}(y)=3(y+44)$
This is usually written in terms of $ \displaystyle x$ so,$ \displaystyle {{g}^{{-1}}}(x)=3(x+44)$